不定積分公式上課講義

1、不定積分公式Ch4、不定積分§1、不定積分的概念與性質(zhì)1、原函數(shù)與不定積分定義1:若F(x)f(x)，則稱F(x)為f(x)的原函數(shù)。 連續(xù)函數(shù)一定有原函數(shù)； 若F(x)為f(x)的原函數(shù)，WJF(x)C也為f(x)的原函數(shù)；事實(shí)上，F(xiàn)(x)C'F'(x)f(x) f(x)的任意兩個(gè)原函數(shù)僅相差一個(gè)常數(shù)。'事實(shí)上，由Fi(x)Fi(x)Fi(x)F2(x)f(x)f(x)0,得Fi(x)F?(x)C故F(x)C表示了f(x)的所有原函數(shù)，其中F(x)為f(x)的一個(gè)原函數(shù)。定義2:f(x)的所有原函數(shù)稱為f(x)的不定積分，記為f(x)dx,積分號,f(x)被

2、積函數(shù)，x積分變量。顯然f(x)dxF(x)CkdxkxC1xxdx12、基本積分表(共24個(gè)基本積分公式)3、不定積分的性質(zhì)f(x)g(x)dxf(x)dxg(x)dxkf(x)dxkf(x)dx(k0)cscxcscxcotxdxcsc2xdxcscxcotxdxcotxcscxCdx.-2sinx22sinxcosx2cosx,22dxsinxcosx2,2cscxdxsecxdxcotxtanxC1、2、cot2xdxcsc2x1dxcotxxC第一類換元法faxbdx1、求不定積分sin5xdx君12x7dxdx,a21dx例2、求不定積分Dx.12.xdx§2、不定積分的

3、換元法（湊微分法）faxbdax1-daxbasin5xd5x5xu1sinudu=52x7d(12x)dxanx1xa2dxn,Lrctanaa一-xarcsina即xn1dxdxn1，、-cos(5x)C512x162x8C(20)(23)2x3,1x3J31x3xedx-edx-e331 1.2 cosdx1.cos-d1.1sinx2xxxxcosx,dx2cosxdx2sin、x2CC3ICdxxc1，x，x,，3、dxdlnx,edxde,sinxdxdcosx,x1dxx2dxcosxdxdsin2x,secxdxdtanx,secxtanxdxdsecx,1x12dxdarct

4、anx,1dxdarcsinx,.a2dx2xda2tanxdxdxIncosxCInsecxC(16)cosxcosxcosxdsinxcotxdxdxInsinxCIncosxC(17)sinxsinxsecxsecxtanxdsecxtanxsecxdxdxInsecxtanxC(18)secxtanxsecxtanxcscxcscxcotx,dcscxcotxcscxdxdxIncscxcotxC(19)sinxdcosx23IcotxcotxcscxcscxdInxInInxCdxxInxInxdx2cosx1dtanx1Intanxtanxtanxxe4x1edxxexeIn1dx

5、x1exex1eIn1exCxeL271edxdex2arctanexxTx2edx1x2d1x1x2例4、求不定積分2x2a2axaxa上ln2axaCxa2xx2.2x1x312xdx12xdx1112dxdx12ad(xxa)adxdx21x21dx1x2【ln2d(xa)xax23arctan(21)(22)二、sin2xdxdx5sin5xcos3xdxcotx切dxInsinxdx71sinxdx8cosxsinx2x26dx52x1lnx2cos2xdx212x2一-dx2x5x22x3一arctan2dxx1211111xcos2xd2xx-sin2xC222242sin2xd

6、x1ocos8x1-cos2xC4sin8x2cosxdxsinInsinx1sinx,2dxcosx第二類換元法1、三角代換例1、Va2x2dxdsinx16dInsinsinxlnsinxdcosx2cosxsec2xdxdx.2sinx4cscxInsinxtanxInInsinxCcosxcscx一4cot解：令xasint（或acost）.a2x2acost,dxacostdt原式=acostacostdt21cos2tadtdtcos2td2t2sin2tC2a一xarcsin2a42a42x22、t212-a2一xarcsinax.a2x2Cdx.xarcsina解：令xasin

7、tacostdtacost-x八dttCarcsinCa例3、dx解：令xatant(或acott)，貝U偵a2x22asect,dxasectdtasec2tdt原式=sectdtinsectasecttantinx2a2Inxx2a2C(24)例*4、解：令xatant(或acott),則x242sect,dx2一2sectdt原式=asec2tdtsectdtasectInsecttantCIn例5、dxHa2解：令xasect(或acsct)x2a2atant,dxasecttantdtasecttantdt原式=sectdtInsect-xtantCIn-a,x2a2c(25)inx

8、x2a2C,-x29例6、dxx解：令xasect，貝UJx293tant,dx3secttantdt2tantdt2,sect13tant.x2933arccosxCx2933arccosCx.a22xxasint小結(jié)：f(x)中含有Jx2a2可考慮用代換xatantx22axasect2、無理代換例7、dx13x1解：令3X1t,則xt31,dx3t2dt23t2dt1t311dt1t1t3t1dt3tIn1t1t2例8、x2133x13ln13x1dxx13x解：令Vxt,則xt6,dx6t5dt5日扯6t5dt原式=飛2t31t2-ydtt2dt6tarctant66xarctan6x

9、me11x_,例9、fdxt,則x解：令1xdxx,xt22tdt21原式=t22tdt"22t1t21 2dtt21產(chǎn)1dt,t1cInCt1例10、2.1x、xln1xxdx解：令瑚1ext,則xInt21,dx2tdtt2112t原式；-dt1dtln2ln/iZ1.一1ex14、倒代換dx例11、6xx，一.1解：令x-tt7原式t6dt124dt1ln4±ln24分部積分公式:UVUVdxd4t614t6124ln4t66x-6x§3、分部積分法UV,UVUVUVdxUVdx,故UdVUVVdU（前后相乘）（前后交換）xdsinxxsinxsinxdxx

10、sinxcosxC例2、xexdxxxxxdexee例3、lnxdxxlndxxexxxdlnexCxxlnx1xdxxln或解：令lnxt,x原式tdettettettedtteetCxxlnxxCxxC例1、xcosxdx例4、arcsinxdxxdx1x2.1x2C§4、兩種典型積分1 xarcsinxxdarcsinxxarcsinxd1x2xarcsinxxarcsinx.1x2或解：令arcsinxt,xsint2原式tdsinttsintsintdttsintcostCxarcsinx,1xC例5、exsinxdxXX.xxXsinxdeesinxecosxdxesin

11、xcosxdexxesinxecosxxedcosxx_.esinxcosxexsinxdx故exsinxdx1exsinxcosxC2x例6、2dxcosxxdtanxxtanxtanxdxxtanxlnsecxC例7、lnxV1x2dx221xxlnx1xxxxlnx.1x21x21x2xdxxlnx.1xdx1x21x2C、有理函數(shù)的積分有理函數(shù)R(x)P(x)Q(x)nanxmn1anixmmbmxbmixa1xa0Kxb0可用待定系數(shù)法化為部分分式，然后積分。例1、將Xx化為部分分式，并計(jì)算%hdx解：-x3x5x6ABx3A2BA3AB12Bx3dxdx故dx565ln(x2)6l

12、n(x3)C1 x5x6x2x3或解：I2x511以1dx25x611dxx25x62x25x62x25x6I2-Inx5x621121 1dxx2211,x3八Inx5x6InC2x2例2、x1xx(x1)2dx1x(x1)1(x1)2dx2dx(x1)2Inx111例3、2x4x1dx2-dx121x2xxx121xx1arctan1xx2dx12x12x11x4124x1dx2例4、dx對三角函數(shù)有理式積分IRsinx,cosxdx,令utan-,2貝Ux2arctanu,2u1u2sinx,cosx,dx1u1u2.du，故I1u1u2Ru2,1u221u2du,三角1dx1xdx1x111x-arctanx,2<21lL2、2x1x2C2x122x122212xxx12x11,2x12xC2.2arctanln2x、.2x2x212xx2二、三角函數(shù)有理式的積分函數(shù)有理式積分即變成了有理函數(shù)積分例5、-3dx5cosx解：令uxtan-,貝Ux2arctanu2cosx1u2,dxudu原式-2.a2du1u1u35-u2duXm222x.2tanln242Cxtan2例6、dx2sinxcosx5解：令uxtan,貝Ux2arctan